3.437 \(\int \frac{(A+B x) \sqrt{a+c x^2}}{\sqrt{e x}} \, dx\)

Optimal. Leaf size=297 \[ \frac{2 a^{3/4} \sqrt{x} \left (\sqrt{a}+\sqrt{c} x\right ) \sqrt{\frac{a+c x^2}{\left (\sqrt{a}+\sqrt{c} x\right )^2}} \left (3 \sqrt{a} B+5 A \sqrt{c}\right ) \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{a}}\right ),\frac{1}{2}\right )}{15 c^{3/4} \sqrt{e x} \sqrt{a+c x^2}}-\frac{4 a^{5/4} B \sqrt{x} \left (\sqrt{a}+\sqrt{c} x\right ) \sqrt{\frac{a+c x^2}{\left (\sqrt{a}+\sqrt{c} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{5 c^{3/4} \sqrt{e x} \sqrt{a+c x^2}}+\frac{2 \sqrt{e x} \sqrt{a+c x^2} (5 A+3 B x)}{15 e}+\frac{4 a B x \sqrt{a+c x^2}}{5 \sqrt{c} \sqrt{e x} \left (\sqrt{a}+\sqrt{c} x\right )} \]

[Out]

(2*Sqrt[e*x]*(5*A + 3*B*x)*Sqrt[a + c*x^2])/(15*e) + (4*a*B*x*Sqrt[a + c*x^2])/(5*Sqrt[c]*Sqrt[e*x]*(Sqrt[a] +
Sqrt[c]*x)) - (4*a^(5/4)*B*Sqrt[x]*(Sqrt[a] + Sqrt[c]*x)*Sqrt[(a + c*x^2)/(Sqrt[a] + Sqrt[c]*x)^2]*EllipticE[
2*ArcTan[(c^(1/4)*Sqrt[x])/a^(1/4)], 1/2])/(5*c^(3/4)*Sqrt[e*x]*Sqrt[a + c*x^2]) + (2*a^(3/4)*(3*Sqrt[a]*B + 5
*A*Sqrt[c])*Sqrt[x]*(Sqrt[a] + Sqrt[c]*x)*Sqrt[(a + c*x^2)/(Sqrt[a] + Sqrt[c]*x)^2]*EllipticF[2*ArcTan[(c^(1/4
)*Sqrt[x])/a^(1/4)], 1/2])/(15*c^(3/4)*Sqrt[e*x]*Sqrt[a + c*x^2])

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Rubi [A]  time = 0.271838, antiderivative size = 297, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {815, 842, 840, 1198, 220, 1196} \[ \frac{2 a^{3/4} \sqrt{x} \left (\sqrt{a}+\sqrt{c} x\right ) \sqrt{\frac{a+c x^2}{\left (\sqrt{a}+\sqrt{c} x\right )^2}} \left (3 \sqrt{a} B+5 A \sqrt{c}\right ) F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{15 c^{3/4} \sqrt{e x} \sqrt{a+c x^2}}-\frac{4 a^{5/4} B \sqrt{x} \left (\sqrt{a}+\sqrt{c} x\right ) \sqrt{\frac{a+c x^2}{\left (\sqrt{a}+\sqrt{c} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{5 c^{3/4} \sqrt{e x} \sqrt{a+c x^2}}+\frac{2 \sqrt{e x} \sqrt{a+c x^2} (5 A+3 B x)}{15 e}+\frac{4 a B x \sqrt{a+c x^2}}{5 \sqrt{c} \sqrt{e x} \left (\sqrt{a}+\sqrt{c} x\right )} \]

Antiderivative was successfully verified.

[In]

Int[((A + B*x)*Sqrt[a + c*x^2])/Sqrt[e*x],x]

[Out]

(2*Sqrt[e*x]*(5*A + 3*B*x)*Sqrt[a + c*x^2])/(15*e) + (4*a*B*x*Sqrt[a + c*x^2])/(5*Sqrt[c]*Sqrt[e*x]*(Sqrt[a] +
Sqrt[c]*x)) - (4*a^(5/4)*B*Sqrt[x]*(Sqrt[a] + Sqrt[c]*x)*Sqrt[(a + c*x^2)/(Sqrt[a] + Sqrt[c]*x)^2]*EllipticE[
2*ArcTan[(c^(1/4)*Sqrt[x])/a^(1/4)], 1/2])/(5*c^(3/4)*Sqrt[e*x]*Sqrt[a + c*x^2]) + (2*a^(3/4)*(3*Sqrt[a]*B + 5
*A*Sqrt[c])*Sqrt[x]*(Sqrt[a] + Sqrt[c]*x)*Sqrt[(a + c*x^2)/(Sqrt[a] + Sqrt[c]*x)^2]*EllipticF[2*ArcTan[(c^(1/4
)*Sqrt[x])/a^(1/4)], 1/2])/(15*c^(3/4)*Sqrt[e*x]*Sqrt[a + c*x^2])

Rule 815

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[((d + e*x)^(
m + 1)*(c*e*f*(m + 2*p + 2) - g*c*d*(2*p + 1) + g*c*e*(m + 2*p + 1)*x)*(a + c*x^2)^p)/(c*e^2*(m + 2*p + 1)*(m
+ 2*p + 2)), x] + Dist[(2*p)/(c*e^2*(m + 2*p + 1)*(m + 2*p + 2)), Int[(d + e*x)^m*(a + c*x^2)^(p - 1)*Simp[f*a
*c*e^2*(m + 2*p + 2) + a*c*d*e*g*m - (c^2*f*d*e*(m + 2*p + 2) - g*(c^2*d^2*(2*p + 1) + a*c*e^2*(m + 2*p + 1)))
*x, x], x], x] /; FreeQ[{a, c, d, e, f, g, m}, x] && NeQ[c*d^2 + a*e^2, 0] && GtQ[p, 0] && (IntegerQ[p] ||  !R
ationalQ[m] || (GeQ[m, -1] && LtQ[m, 0])) &&  !ILtQ[m + 2*p, 0] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*
m, 2*p])

Rule 842

Int[((f_) + (g_.)*(x_))/(Sqrt[(e_)*(x_)]*Sqrt[(a_) + (c_.)*(x_)^2]), x_Symbol] :> Dist[Sqrt[x]/Sqrt[e*x], Int[
(f + g*x)/(Sqrt[x]*Sqrt[a + c*x^2]), x], x] /; FreeQ[{a, c, e, f, g}, x]

Rule 840

Int[((f_) + (g_.)*(x_))/(Sqrt[x_]*Sqrt[(a_) + (c_.)*(x_)^2]), x_Symbol] :> Dist[2, Subst[Int[(f + g*x^2)/Sqrt[
a + c*x^4], x], x, Sqrt[x]], x] /; FreeQ[{a, c, f, g}, x]

Rule 1198

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[c/a, 2]}, Dist[(e + d*q)/q, Int
[1/Sqrt[a + c*x^4], x], x] - Dist[e/q, Int[(1 - q*x^2)/Sqrt[a + c*x^4], x], x] /; NeQ[e + d*q, 0]] /; FreeQ[{a
, c, d, e}, x] && PosQ[c/a]

Rule 220

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 4]}, Simp[((1 + q^2*x^2)*Sqrt[(a + b*x^4)/(a*(
1 + q^2*x^2)^2)]*EllipticF[2*ArcTan[q*x], 1/2])/(2*q*Sqrt[a + b*x^4]), x]] /; FreeQ[{a, b}, x] && PosQ[b/a]

Rule 1196

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[c/a, 4]}, -Simp[(d*x*Sqrt[a + c
*x^4])/(a*(1 + q^2*x^2)), x] + Simp[(d*(1 + q^2*x^2)*Sqrt[(a + c*x^4)/(a*(1 + q^2*x^2)^2)]*EllipticE[2*ArcTan[
q*x], 1/2])/(q*Sqrt[a + c*x^4]), x] /; EqQ[e + d*q^2, 0]] /; FreeQ[{a, c, d, e}, x] && PosQ[c/a]

Rubi steps

\begin{align*} \int \frac{(A+B x) \sqrt{a+c x^2}}{\sqrt{e x}} \, dx &=\frac{2 \sqrt{e x} (5 A+3 B x) \sqrt{a+c x^2}}{15 e}+\frac{4 \int \frac{\frac{5}{2} a A c e^2+\frac{3}{2} a B c e^2 x}{\sqrt{e x} \sqrt{a+c x^2}} \, dx}{15 c e^2}\\ &=\frac{2 \sqrt{e x} (5 A+3 B x) \sqrt{a+c x^2}}{15 e}+\frac{\left (4 \sqrt{x}\right ) \int \frac{\frac{5}{2} a A c e^2+\frac{3}{2} a B c e^2 x}{\sqrt{x} \sqrt{a+c x^2}} \, dx}{15 c e^2 \sqrt{e x}}\\ &=\frac{2 \sqrt{e x} (5 A+3 B x) \sqrt{a+c x^2}}{15 e}+\frac{\left (8 \sqrt{x}\right ) \operatorname{Subst}\left (\int \frac{\frac{5}{2} a A c e^2+\frac{3}{2} a B c e^2 x^2}{\sqrt{a+c x^4}} \, dx,x,\sqrt{x}\right )}{15 c e^2 \sqrt{e x}}\\ &=\frac{2 \sqrt{e x} (5 A+3 B x) \sqrt{a+c x^2}}{15 e}-\frac{\left (4 a^{3/2} B \sqrt{x}\right ) \operatorname{Subst}\left (\int \frac{1-\frac{\sqrt{c} x^2}{\sqrt{a}}}{\sqrt{a+c x^4}} \, dx,x,\sqrt{x}\right )}{5 \sqrt{c} \sqrt{e x}}+\frac{\left (4 a \left (3 \sqrt{a} B+5 A \sqrt{c}\right ) \sqrt{x}\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+c x^4}} \, dx,x,\sqrt{x}\right )}{15 \sqrt{c} \sqrt{e x}}\\ &=\frac{2 \sqrt{e x} (5 A+3 B x) \sqrt{a+c x^2}}{15 e}+\frac{4 a B x \sqrt{a+c x^2}}{5 \sqrt{c} \sqrt{e x} \left (\sqrt{a}+\sqrt{c} x\right )}-\frac{4 a^{5/4} B \sqrt{x} \left (\sqrt{a}+\sqrt{c} x\right ) \sqrt{\frac{a+c x^2}{\left (\sqrt{a}+\sqrt{c} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{5 c^{3/4} \sqrt{e x} \sqrt{a+c x^2}}+\frac{2 a^{3/4} \left (3 \sqrt{a} B+5 A \sqrt{c}\right ) \sqrt{x} \left (\sqrt{a}+\sqrt{c} x\right ) \sqrt{\frac{a+c x^2}{\left (\sqrt{a}+\sqrt{c} x\right )^2}} F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{15 c^{3/4} \sqrt{e x} \sqrt{a+c x^2}}\\ \end{align*}

Mathematica [C]  time = 0.0263668, size = 82, normalized size = 0.28 \[ \frac{2 x \sqrt{a+c x^2} \left (3 A \, _2F_1\left (-\frac{1}{2},\frac{1}{4};\frac{5}{4};-\frac{c x^2}{a}\right )+B x \, _2F_1\left (-\frac{1}{2},\frac{3}{4};\frac{7}{4};-\frac{c x^2}{a}\right )\right )}{3 \sqrt{e x} \sqrt{\frac{c x^2}{a}+1}} \]

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x)*Sqrt[a + c*x^2])/Sqrt[e*x],x]

[Out]

(2*x*Sqrt[a + c*x^2]*(3*A*Hypergeometric2F1[-1/2, 1/4, 5/4, -((c*x^2)/a)] + B*x*Hypergeometric2F1[-1/2, 3/4, 7
/4, -((c*x^2)/a)]))/(3*Sqrt[e*x]*Sqrt[1 + (c*x^2)/a])

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Maple [A]  time = 0.023, size = 312, normalized size = 1.1 \begin{align*}{\frac{2}{15\,c} \left ( 5\,A\sqrt{-ac}\sqrt{{\frac{cx+\sqrt{-ac}}{\sqrt{-ac}}}}\sqrt{2}\sqrt{{\frac{-cx+\sqrt{-ac}}{\sqrt{-ac}}}}\sqrt{-{\frac{cx}{\sqrt{-ac}}}}{\it EllipticF} \left ( \sqrt{{\frac{cx+\sqrt{-ac}}{\sqrt{-ac}}}},1/2\,\sqrt{2} \right ) a+6\,B\sqrt{{\frac{cx+\sqrt{-ac}}{\sqrt{-ac}}}}\sqrt{2}\sqrt{{\frac{-cx+\sqrt{-ac}}{\sqrt{-ac}}}}\sqrt{-{\frac{cx}{\sqrt{-ac}}}}{\it EllipticE} \left ( \sqrt{{\frac{cx+\sqrt{-ac}}{\sqrt{-ac}}}},1/2\,\sqrt{2} \right ){a}^{2}-3\,B\sqrt{{\frac{cx+\sqrt{-ac}}{\sqrt{-ac}}}}\sqrt{2}\sqrt{{\frac{-cx+\sqrt{-ac}}{\sqrt{-ac}}}}\sqrt{-{\frac{cx}{\sqrt{-ac}}}}{\it EllipticF} \left ( \sqrt{{\frac{cx+\sqrt{-ac}}{\sqrt{-ac}}}},1/2\,\sqrt{2} \right ){a}^{2}+3\,B{c}^{2}{x}^{4}+5\,A{c}^{2}{x}^{3}+3\,aBc{x}^{2}+5\,aAcx \right ){\frac{1}{\sqrt{c{x}^{2}+a}}}{\frac{1}{\sqrt{ex}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)*(c*x^2+a)^(1/2)/(e*x)^(1/2),x)

[Out]

2/15/(c*x^2+a)^(1/2)/c*(5*A*(-a*c)^(1/2)*((c*x+(-a*c)^(1/2))/(-a*c)^(1/2))^(1/2)*2^(1/2)*((-c*x+(-a*c)^(1/2))/
(-a*c)^(1/2))^(1/2)*(-x*c/(-a*c)^(1/2))^(1/2)*EllipticF(((c*x+(-a*c)^(1/2))/(-a*c)^(1/2))^(1/2),1/2*2^(1/2))*a
+6*B*((c*x+(-a*c)^(1/2))/(-a*c)^(1/2))^(1/2)*2^(1/2)*((-c*x+(-a*c)^(1/2))/(-a*c)^(1/2))^(1/2)*(-x*c/(-a*c)^(1/
2))^(1/2)*EllipticE(((c*x+(-a*c)^(1/2))/(-a*c)^(1/2))^(1/2),1/2*2^(1/2))*a^2-3*B*((c*x+(-a*c)^(1/2))/(-a*c)^(1
/2))^(1/2)*2^(1/2)*((-c*x+(-a*c)^(1/2))/(-a*c)^(1/2))^(1/2)*(-x*c/(-a*c)^(1/2))^(1/2)*EllipticF(((c*x+(-a*c)^(
1/2))/(-a*c)^(1/2))^(1/2),1/2*2^(1/2))*a^2+3*B*c^2*x^4+5*A*c^2*x^3+3*a*B*c*x^2+5*a*A*c*x)/(e*x)^(1/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{c x^{2} + a}{\left (B x + A\right )}}{\sqrt{e x}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+a)^(1/2)/(e*x)^(1/2),x, algorithm="maxima")

[Out]

integrate(sqrt(c*x^2 + a)*(B*x + A)/sqrt(e*x), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{c x^{2} + a}{\left (B x + A\right )} \sqrt{e x}}{e x}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+a)^(1/2)/(e*x)^(1/2),x, algorithm="fricas")

[Out]

integral(sqrt(c*x^2 + a)*(B*x + A)*sqrt(e*x)/(e*x), x)

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Sympy [C]  time = 3.39806, size = 97, normalized size = 0.33 \begin{align*} \frac{A \sqrt{a} \sqrt{x} \Gamma \left (\frac{1}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{1}{2}, \frac{1}{4} \\ \frac{5}{4} \end{matrix}\middle |{\frac{c x^{2} e^{i \pi }}{a}} \right )}}{2 \sqrt{e} \Gamma \left (\frac{5}{4}\right )} + \frac{B \sqrt{a} x^{\frac{3}{2}} \Gamma \left (\frac{3}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{1}{2}, \frac{3}{4} \\ \frac{7}{4} \end{matrix}\middle |{\frac{c x^{2} e^{i \pi }}{a}} \right )}}{2 \sqrt{e} \Gamma \left (\frac{7}{4}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x**2+a)**(1/2)/(e*x)**(1/2),x)

[Out]

A*sqrt(a)*sqrt(x)*gamma(1/4)*hyper((-1/2, 1/4), (5/4,), c*x**2*exp_polar(I*pi)/a)/(2*sqrt(e)*gamma(5/4)) + B*s
qrt(a)*x**(3/2)*gamma(3/4)*hyper((-1/2, 3/4), (7/4,), c*x**2*exp_polar(I*pi)/a)/(2*sqrt(e)*gamma(7/4))

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{c x^{2} + a}{\left (B x + A\right )}}{\sqrt{e x}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+a)^(1/2)/(e*x)^(1/2),x, algorithm="giac")

[Out]

integrate(sqrt(c*x^2 + a)*(B*x + A)/sqrt(e*x), x)